Solution:
Let E1 be the event that Both A and B solve the problem.
A and B are independent,
$\Rightarrow\ \text{P}(\text{E}_1)=\text{P(A)}\times\text{P(B)}$
$\Rightarrow\ \text{P}(\text{E}_1)=\frac{1}{3}\times\frac{1}{4}=\frac{1}{12}$
Let E2 both A and B got wrong solution.
$\text{P}(\text{E}_2)=\Big(1-\frac{1}{3}\Big)\times\Big(1-\frac{1}{4}\Big)=\frac{1}{2}$
Let E be the event getting same answer.
$\text{P}\Big(\frac{\text{E}}{\text{E}_1}\Big)=1,\text{P}\Big(\frac{\text{E}}{\text{E}_2}\Big)=\frac{1}{20}$
$\Rightarrow\text{P}\Big(\frac{\text{E}_1}{\text{E}}\Big)=\frac{\frac{1}{12}\times1}{\frac{1}{12}\times1+\frac{1}{2}\times\frac{1}{20}}=\frac{10}{13}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$2\int\limits^{\text{a}}_0\text{f(x)}\text{dx}$
$0$
$\int\limits^{\text{a}}_0\text{f}(\text{x})\text{dx}+\int\limits^{\text{a}}_0\text{f}(2\text{a}-\text{x})\text{dx}$
$\int\limits^{\text{a}}_0\text{f}(\text{x})\text{dx}+\int\limits^{2\text{a}}_0\text{f}(2\text{a}-\text{x})\text{dx}$